Answer:
2630250 N/C, horizontally left
0
[tex]1.67\times 10^{-8}\ s[/tex]
1434.825 N/C, horizontally left
Explanation:
m = Mass of particle
u = Initial velocity = [tex]4.2\times 10^6\ m/s[/tex]
v = Final velocity = 0
t = Time taken
s = Displacement = 3.5 cm
q = Charge of particle = [tex]1.6\times 10^{-19}\ C[/tex]
Force is given by
[tex]F=qE[/tex]
Acceleration is given by
[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times E}{1.67\times 10^{-27}}\\\Rightarrow a=95808383.23353E[/tex]
[tex]v^2-u^2=2as\\\Rightarrow v^2-u^2=2\times 95808383.23353Es\\\Rightarrow E=\dfrac{0^2-(4.2\times 10^6)^2}{2\times 95808383.23353\times 0.035}\\\Rightarrow E=-2630250\ N/C[/tex]
Magnitude of electric field is 2630250 N/C
Direction is horizontally to the left
The angle counterclockwise from left is zero.
[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times -2630250}{1.67\times 10^{-27}}\\\Rightarrow a=-2.52\times 10^{14}\ m/s^2[/tex]
[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-4.2\times 10^6}{-2.52\times 10^{14}}\\\Rightarrow t=1.67\times 10^{-8}\ s[/tex]
The time taken is [tex]1.67\times 10^{-8}\ s[/tex]
Acceleration is given by
[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times E}{9.11\times 10^{-31}}\\\Rightarrow a=175631174533.4797E[/tex]
[tex]v^2-u^2=2as\\\Rightarrow v^2-u^2=2\times 175631174533.4797Es\\\Rightarrow E=\dfrac{0^2-(4.2\times 10^6)^2}{2\times 175631174533.4797\times 0.035}\\\Rightarrow E=-1434.825\ N/C[/tex]
Magnitude of electric field is 1434.825 N/C
Direction is horizontally to the left
